InterviewSolution
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InterviewSolution
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A monatomic idea gas of `2 mol` is taken through a cyclic process starting from `A` as shown in figure. The volume ratio are `V_(B)//V_(A)=2` and `V_(D)//V_(A)=4`. If the temperature `T_(A)` at `A` is `27^(@)C`, and gas constant is `R`. Calculate. The temperature of the gas at point `B`A. `500K`B. `700K`C. `600 K`D. `300 K` |
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Answer» Correct Answer - C Gas is monatomic, hence `C_(v)=(3)/(2)R` and `C_(p)=(5)/(2)R` Number of moles, `n=2` `T_(A)=27^(@)C=300K` Process `A rarr B` is a striaght line passing through origin, hence it is an isobaric process. `(V_(A))/(T_(A))=(V_(B))/(T_(B))` `T_(B)=((V_(B))/(V_(A)))T_(A)` `=2 xx 300 K=600K` ltbr. Porcess `A rarr B ` is isobaric. `Q_(p)=nC_(p)(T_(B)-T_(A))` `Q_(p)=2((5)/( 2)R)(600-300)=1500R` Heat is added to the system. Process` B rarr C` is isothermal. From the first law of thermodynamics. `Q_(BC)=W_(BC)` `(` as `Delta U=0)` ` W=nRT_(B)1n((V_(C))/(V_(B)))` `=(2)(R)(600)1n((4V_(A))/(2V_(A)))` `=(1200R)1n (2)` Heat is added to the system. Process `C rarr D ` is isochoric. From the first law of thermodynamics, `Q_(CD)=nC_(v)(T_(D)-T_(C))=n((3)/(2)R)(T_(A)-T_(B))` `=2((3)/(2) R)(300-600)=-900R` Negative sign implies that heat is rejected by the system. Process `D rarr A` is isothermal. `Q_(DA)=W_(DA)=nRT_(0)1n((V_(A))/(V_(D)))` `=(2)(R)3001n((V_(A))/(4V_(A)))=600R 1n((1)/(4))` Negative sign implies that heat is rejected by the system. In a cyclic process, `Delta U=0` `Q_(n et)=W_( n et)` `W_(n et)=Q_(AB)+Q_(BC)+Q_(CD)+Q_(DA)` `=1500 R+8.316R-900R-831.6R=600R` |
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